# Pumping Lemma - unregular expression

How do prove that this expression is unregular, I know firstly you have to try prove that it is regular and work from there. I also know that $$w=xuz$$ and the three rules are needed

Let $$M$$ be the language over the alphabet $$\{a, b, c\}$$ given by $$M = \{a^ib^jc^k \mid i, j, k ≥ 0, j = i + k\}$$.

• What you provide is called a "language", not an "expression". Also "unregular" is a very uncommon term; usually one says "not regular" or "non-regular". – Peter Leupold Jan 24 at 12:15

Suppose $$M$$ is regular. Then by the Pumping Lemma, there is some $$p$$ so that all words of length at least $$p$$ can be decomposed as $$xyz$$ in such a way that $$xy^nz\in M$$ for all $$n$$.
Consider the word $$ab^pc^{p-1}\in M$$, and say that its Pumping Lemma decomposition is $$xyz$$. What are $$x$$, $$y$$, and $$z$$?
Prove that if $$xy^nz\in M$$ for all $$n\in\mathbb{N}$$, then necessarily a few properties must hold:
1. $$x$$ must contain $$a$$ and cannot contain any $$c$$
2. $$z$$ must contain all $$p-1$$ copies of $$c$$
This necessarily means that we have $$x=ab^{m_x}$$, $$y=b^{m_y}$$, and $$z=b^{m_z}c^{p-1}$$ for some $$m_x,m_y,m_z\geq 0$$ such that $$m_x+m_y+m_z=p$$ and $$m_y\geq 1$$. But then $$xy^nz=ab^{m_x+nm_y+m_z}c^{p-1}=ab^{p+(n-1)m_y}c^{p-1}.$$ But, this word clearly cannot be in $$M$$, as $$p+(n-1)m_y>1+(p-1)$$ for all $$n\geq 2$$.